n-1-n-2n-1-help-me-pls-

Question Number 100468 by Mikael_786 last updated on 26/Jun/20

\underset{n = 1}{\sum^{\infty}} \frac{n}{(2 n + 1)!}

h e l p m e p l s

Answered by mathmax by abdo last updated on 26/Jun/20

S = \frac{1}{2} \sum_{n = 0}^{\infty} \frac{2 n + 1 - 1}{(2 n + 1)!} = \frac{1}{2} \sum_{n = 0}^{\infty} \frac{1}{(2 n)!} - \frac{1}{2} \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)!}

we have e^{x} + e^{- x} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} + \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{n}}{n!} = \sum_{n = 0}^{\infty} \frac{1}{n!} (1 + {(- 1)}^{n}) x^{n}

= 2 \sum_{n = 0}^{\infty} \frac{x^{2 n}}{(2 n)!} \Rightarrow \frac{e^{x} + e^{- x}}{2} = \sum_{n = 0}^{\infty} \frac{x^{2 n}}{(2 n)!} \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{(2 n)!} = \frac{e + e^{- 1}}{2}

e^{x} - e^{- x} = \sum_{n = 0}^{\infty} \frac{1}{n!} (1 - {(- 1)}^{n}) x^{n} = 2 \sum_{n = 0}^{\infty} \frac{x^{2 n + 1}}{(2 n + 1)!} \Rightarrow

\sum_{n = 0}^{\infty} \frac{x^{2 n + 1}}{(2 n + 1)!} = \frac{e^{x} - e^{- x}}{2} \Rightarrow \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)!} = \frac{e - e^{- 1}}{2} \Rightarrow

S = \frac{1}{4} (e + e^{- 1}) - \frac{1}{4} (e - e^{- 1}) = \frac{1}{4} (e + e^{- 1} - e + e^{- 1}) = \frac{{2 e}^{- 1}}{4} = \frac{1}{2 e}

S = \frac{1}{2 e}

Commented by Mikael_786 last updated on 27/Jun/20

t h a n k y o u S i r

Answered by smridha last updated on 26/Jun/20

\boldsymbol a n s : \frac{1}{2 \boldsymbol e} . \boldsymbol a p p l y \boldsymbol g e n e r a l i s e d

\boldsymbol h y p e r g e o m e t r i c \boldsymbol f^{\boldsymbol n}

Answered by maths mind last updated on 26/Jun/20

f (x) = \sum_{k ⩾ 1} \frac{x^{2 n + 1}}{(2 n + 1)!} = \frac{e^{x} - e^{- x}}{2}

\Rightarrow \frac{e^{x} - e^{- x}}{2 x} = \sum_{n ⩾ 1} \frac{x^{2 n}}{(2 n + 1)!}

\partial x \frac{1}{2} (\frac{e^{x} - e^{- x}}{x}) ∣_{x = 1} = \sum_{k ⩾ 1} \frac{2 n}{(2 n + 1)!}

\Leftrightarrow \frac{1}{z} (\frac{e^{x} + e^{- x}}{x} - \frac{e^{x} - e^{- x}}{x^{2}}) ∣_{x = 1} = 2 \underset{n = 1}{\sum^{+ \infty}} \frac{n}{(2 n + 1)!}

\Leftrightarrow e^{- 1} = 2 Σ \frac{n}{(2 n + 1)!} \Rightarrow \frac{1}{2 e} = \sum_{n ⩾ 1} \frac{n}{(2 n + 1)!}

Commented by Mikael_786 last updated on 27/Jun/20

t h a n k y o u S i r

t h a n k y o u S i r